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Andru [333]
3 years ago
15

A survey of two communities asked residents which candidate they supported for a local election. The survey data are shown in th

e relative frequency table.
What percentage of the Mountain View residents polled supported Olunloyo?

A. 18%
B. 30%
C. 37.5%
D. About 47%​

Mathematics
1 answer:
pashok25 [27]3 years ago
7 0

Answer:

the answer is D

Step-by-step explanation:

i believe it because of the total

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For the people who helped me with the question
erik [133]

Answer:

glad my answer helped you.

Step-by-step explanation:

hope that because i showed the steps, it will help you if you come across a problem like that in the future too!

4 0
3 years ago
Fraction form or decimal form! PLEASE HELP
joja [24]

Answer:

x = undefined

Step-by-step explanation:

Because you have an x and then remove the x the it is not pssible to tell what x was with the given information causing it to be undefined.

3 0
3 years ago
If 9% of the bar’s weight is iron, how much does the iron weigh
Leya [2.2K]

Answer:

15 pounds

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
At a local Dunkin’ Donuts, a survey showed that out of 1,300 customers eating lunch, 520 ordered coffee with their meal. What pe
k0ka [10]

First subtract the number that did, to find the number that didn't.

1300 - 520 = 780 customers did not order coffee.

Now divide the number who didn't order coffee by total number of customers:

780 / 1300 = 0.6

Multiply by 100 to get the percent:

0.6 x 100 = 60% did not order coffee.

5 0
3 years ago
The life span of a certain insect​ (in days) is uniformly distributed over the interval left bracket 15.29 right bracket . A. Wh
Leni [432]

The exercise says that the life variable is measured in days, that is, it is a discrete distribution, this because it takes finite and punctual values, 21, 22, 23 days, etc., not 21 and a half days or 23 and a quarter days.

Now, the discrete uniform distribution assumes that the probability of each value it takes is equal.

In this case the distribution takes 15 values, from 15 to 29 including the extremes, so the probability of each value is \frac{1}{n} = \frac{1}{15}, so the probability mass function will be:

f(x)=\left \{ {{\frac{1}{15},x=15,16,..,28,29} \atop {0, else}} \right.

With this we can calculate the probability that an insect, selected at random, lives more than 24 days, like this:

P (x> 24) = f (25) + f (26) + f (27) + f (28) + f (29) + f (30) + f (31) ...\\P (x> 24) = \frac{1}{15} + \frac{1}{15} + \frac{1}{15} + \frac{1}{15} + \frac{1}{15} + 0 + 0 + ...\\P (x> 24) = \frac{5}{15} = \frac{1}{3}

On the other hand, the expected life or average life in this distribution is calculated as follows: \frac{a+b}{2} that this case is \frac{15+29}{2} = 22

Answer

A) The awakened life of the insect is 22 days

B) The probability that an insect, selected at random, lives more than 24 days is 1/3 = 0.33 = 33.3%

5 0
3 years ago
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